The Dual Space and Dual Bases

Dual spaces are not a curiosity; they organize huge swaths of mathematics and its applications. In differential geometry, the dual of the tangent space at a point is the cotangent space, whose elements (differential 1-forms like $dx, dy$) are exactly linear functionals in the sense defined here, and the dual basis construction is precisely how one builds coordinate 1-forms from coordinate vector fields. In quantum mechanics, bra vectors $\langle \psi |$ live in the dual of the space of ket vectors $|\psi\rangle$, and the dual pairing $\langle \psi | \phi \rangle$ is the physicist's notation for evaluating a functional on a vector. In statistics and machine learning, linear functionals appear as linear regression weight vectors acting on feature vectors, and dual-basis ideas underlie duality in constrained optimization (Lagrangian duality, support vector machines). In numerical analysis, evaluation functionals like the ones in the polynomial example above are the building blocks of quadrature rules and finite-element methods, where "dual bases" of nodal functionals are used to interpolate functions exactly. In cryptography and coding theory, dual bases of finite field extensions are used to build efficient bit representations and trace-based pairings. Finally, within pure linear algebra itself, the dual space is the essential stepping stone to the dual map (the transpose of a linear map, viewed conceptually) and to the double dual $V'' = (V')'$, where finite-dimensional $V$ reappears canonically inside its own double dual — a phenomenon that only makes sense once the dual basis and the dimension theorem proved here are firmly in hand.

Ask the tutor: How does changing the basis of $V$ transform its dual basis, and why does that transformation involve the inverse transpose of the change-of-basis matrix?

Think of a vector $v \in V$ as a point located by its coordinates relative to a fixed basis $(v_1,\dots,v_n)$: there are unique scalars $a_1,\dots,a_n$ with $v = a_1 v_1 + \cdots + a_n v_n$. The dual basis functional $\varphi_i$ is nothing but the operation "read off $a_i$." Indeed by linearity,

since every term vanishes except the one with $\varphi_i(v_i) = 1$. So $\varphi_i$ is literally the $i$-th coordinate function, and the biorthogonality relation $\varphi_i(v_j) = \delta_{ij}$ is just the statement that reading off the $i$-th coordinate of the $j$-th basis vector gives $1$ if $i=j$ (since $v_j$ has coordinate vector $e_j$) and $0$ otherwise. This gives a second, complementary picture of a basis: instead of $n$ vectors that span and are independent, you can equally picture $n$ "measuring instruments" (linear functionals) that together recover every coordinate of every vector, and are themselves independent as functions. Changing the basis of $V$ changes both pictures in lock-step — a fact that becomes central once one studies the dual map and change-of-basis matrices for $V'$.

Example 1: The dual basis in $\mathbf{F}^3$ relative to a non-standard basis. Let $V = \mathbf{R}^3$ and take the (deliberately non-standard, asymmetric) ordered basis

First check this is really a basis: since $\dim \mathbf{R}^3 = 3$ and we have $3$ vectors, it suffices to check linear independence, e.g. via the determinant

so $(v_1,v_2,v_3)$ is indeed a basis. We seek the dual basis $(\varphi_1,\varphi_2,\varphi_3)$, writing each $\varphi_i(x,y,z) = a_i x + b_i y + c_i z$ for scalars to be determined, and imposing $\varphi_i(v_j) = \delta_{ij}$. For $\varphi_1 = (a,b,c)$ (identifying a functional on $\mathbf{R}^3$ with its coefficient row via the standard dot product pairing), the conditions are

From the second equation $c = -b$. Substituting into the third: $2a - b = 0 \Rightarrow b = 2a$. Substituting into the first: $a + 2a = 1 \Rightarrow a = 1/3$, so $b = 2/3$, $c = -2/3$. Thus

Check: $\varphi_1(v_1) = \tfrac13(1) + \tfrac23(1) - \tfrac23(0) = \tfrac13+\tfrac23 = 1$. ✓ $\varphi_1(v_2) = \tfrac13(0)+\tfrac23(1)-\tfrac23(1) = \tfrac23 - \tfrac23 = 0$. ✓ $\varphi_1(v_3) = \tfrac13(2)+\tfrac23(0)-\tfrac23(1) = \tfrac23 - \tfrac23 = 0$. ✓ Next, $\varphi_2 = (a,b,c)$ satisfies $a+b=0$, $b+c=1$, $2a+c=0$. From the first, $b=-a$; from the third, $c = -2a$. Substitute into the second: $-a - 2a = 1 \Rightarrow a = -1/3$, so $b = 1/3$, $c = 2/3$. Thus

Check: $\varphi_2(v_1) = -\tfrac13+\tfrac13 = 0$ ✓, $\varphi_2(v_2) = \tfrac13+\tfrac23=1$ ✓, $\varphi_2(v_3) = -\tfrac23+\tfrac23=0$ ✓. Finally $\varphi_3 = (a,b,c)$ satisfies $a+b=0$, $b+c=0$, $2a+c=1$. From the first $b=-a$, from the second $c=-b=a$. Substitute into the third: $2a+a=1 \Rightarrow a=1/3$, so $b=-1/3$, $c=1/3$. Thus

Check: $\varphi_3(v_1) = \tfrac13-\tfrac13=0$ ✓, $\varphi_3(v_2)=-\tfrac13+\tfrac13=0$ ✓, $\varphi_3(v_3)=\tfrac23+\tfrac13=1$ ✓. Now test the coordinate formula on a fresh vector, say $u = (5,1,4)$. Solving $u = a v_1 + b v_2 + c v_3$: this is the same linear system with right side $(5,1,4)^T$ against the columns $v_1,v_2,v_3$; using Cramer's rule with the same determinant $3$ from before, one finds (after direct substitution) $a=1$, $b=2$, $c=2$, since $1\cdot(1,1,0) + 2\cdot(0,1,1) + 2\cdot(2,0,1) = (1,1,0)+(0,2,2)+(4,0,2) = (5,3,4)$ — that does not match $(5,1,4)$, so let's instead directly verify via the functionals, which is more robust: $\varphi_1(u) = \tfrac13(5)+\tfrac23(1)-\tfrac23(4) = \tfrac{5+2-8}{3} = -\tfrac13$, $\varphi_2(u) = -\tfrac13(5)+\tfrac13(1)+\tfrac23(4) = \tfrac{-5+1+8}{3} = \tfrac43$, $\varphi_3(u) = \tfrac13(5)-\tfrac13(1)+\tfrac13(4) = \tfrac{5-1+4}{3} = \tfrac83$. As a check of consistency, form $-\tfrac13 v_1 + \tfrac43 v_2 + \tfrac83 v_3 = -\tfrac13(1,1,0) + \tfrac43(0,1,1) + \tfrac83(2,0,1) = (-\tfrac13,-\tfrac13,0)+(0,\tfrac43,\tfrac43)+(\tfrac{16}{3},0,\tfrac83) = \left(\tfrac{-1+16}{3}, \tfrac{-1+4}{3}, \tfrac{4+8}{3}\right) = (5,1,4) = u$. ✓ This confirms $\varphi_i(u)$ really are the coordinates of $u$ in the basis $(v_1,v_2,v_3)$, exactly as the theorem predicts.

Example 2: Dual basis for polynomials via evaluation-type functionals. Let $V = \mathcal{P}_2(\mathbf{R})$, the (3-dimensional) space of real polynomials of degree at most $2$, with the ordered basis

This is a basis because it consists of $3$ nonzero polynomials of strictly increasing degree $0,1,2$, which forces linear independence (a nontrivial linear combination summing to zero would force the top-degree coefficient of the highest-degree term present to vanish, contradiction, by induction on degree), and $3 = \dim \mathcal{P}_2(\mathbf{R})$. We claim the dual basis is given by evaluation functionals at cleverly chosen points. Try

where $\operatorname{ev}_a(p) = p(a)$, and see whether biorthogonality actually holds (it need not automatically — we must check, and adjust if it fails). Compute the $3\times 3$ table of values $\operatorname{ev}_a(p_j)$:

• $p_1(1) = 1$, $p_1(2) = 1$, $p_1(0) = 1$.
• $p_2(1) = 0$, $p_2(2) = 1$, $p_2(0) = -1$.
• $p_3(1) = (0)(-1) = 0$, $p_3(2) = (1)(0) = 0$, $p_3(0) = (-1)(-2) = 2$.

So relative to $(p_1,p_2,p_3)$, the functional $\operatorname{ev}_1$ has values $(1,0,0)$ — a direct hit, since $\operatorname{ev}_1(p_1)=1$, $\operatorname{ev}_1(p_2)=0$, $\operatorname{ev}_1(p_3)=0$. So $\varphi_1 = \operatorname{ev}_1$ works exactly as the first dual basis vector. The functional $\operatorname{ev}_2$ has values $(1,1,0)$: it does not vanish on $p_1$, so $\operatorname{ev}_2$ is not $\varphi_2$ on the nose; we must correct it. We want $\varphi_2$ with $\varphi_2(p_1)=0,\varphi_2(p_2)=1,\varphi_2(p_3)=0$. Try $\varphi_2 = \operatorname{ev}_2 - \operatorname{ev}_1$ (a legitimate element of $V'$, since $V'$ is a vector space closed under linear combinations). Then

This matches exactly the requirement for the second dual basis vector, so $\varphi_2 = \operatorname{ev}_2 - \operatorname{ev}_1$, i.e. $\varphi_2(p) = p(2) - p(1)$. The functional $\operatorname{ev}_0$ has values $(1,-1,2)$. We want $\varphi_3$ with $(0,0,1)$. Using the basis $\{\varphi_1,\varphi_2,\operatorname{ev}_0\}$ of $V'$ (justified because these three functionals inherit linear independence from evaluating at three distinct points on a 3-dimensional space — or simply because we now solve directly), set $\varphi_3 = \operatorname{ev}_0 - \alpha \varphi_1 - \beta\varphi_2$ and solve for $\alpha,\beta$ so that $\varphi_3(p_1)=\varphi_3(p_2) = 0$ while $\varphi_3(p_3)=1$ automatically holds since $\varphi_1(p_3)=\varphi_2(p_3)=0$ already, matching $\operatorname{ev}_0(p_3) = 2 \ne 1$ — so we must also rescale. Concretely:

so $\varphi_3 = \operatorname{ev}_0 - \varphi_1 + \varphi_2$, and then $\varphi_3(p_3) = 2 - 0 + 0 = 2 \ne 1$. Since $V'$ is a vector space, we simply rescale: define instead

Then $\varphi_3(p_1) = 0$, $\varphi_3(p_2) = 0$, $\varphi_3(p_3) = 2/2 = 1$, all as required, so this is the genuine third dual basis vector. Explicitly, unwinding $\varphi_1 = \operatorname{ev}_1$ and $\varphi_2 = \operatorname{ev}_2 - \operatorname{ev}_1$,

Sanity check via the coordinate formula. Take $q(x) = 3 + 2(x-1) - (x-1)(x-2) = 3p_1 + 2p_2 - p_3$, so its coordinates relative to $(p_1,p_2,p_3)$ are $(3,2,-1)$. The theorem predicts $\varphi_1(q)=3$, $\varphi_2(q)=2$, $\varphi_3(q)=-1$. Expand $q(x) = 3 + 2x - 2 - (x^2-3x+2) = 3+2x-2-x^2+3x-2 = -x^2+5x-1$. Then $q(0) = -1$, $q(1) = -1+5-1=3$, $q(2) = -4+10-1=5$. So $\varphi_1(q) = q(1) = 3$ ✓; $\varphi_2(q) = q(2)-q(1) = 5-3 = 2$ ✓; $\varphi_3(q) = \tfrac12 q(0) - q(1) + \tfrac12 q(2) = \tfrac12(-1) - 3 + \tfrac12(5) = -0.5-3+2.5 = -1$ ✓. All three match perfectly, confirming the dual basis construction and the coordinate-reading interpretation.